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QJEEM23Phad06: JEE Main 2023 Physics - Class 12 JEEM23Phad06 Solution: A light of energy is incident on a hydrogen atom in its ground…

A light of energy 12.75 eV12.75 ~\mathrm{eV} is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is xπ×10−17 eVs\frac{x}{\pi} \times 10^{-17} ~\mathrm{eVs} . The value of xx is ___________ (use h=4.14×10−15 eVs,c=3×108 ms−1h=4.14 \times 10^{-15} ~\mathrm{eVs}, c=3 \times 10^{8} \mathrm{~ms}^{-1} ).

Step-by-Step Verified Solution

Let the electron jumps to nth n^{\text {th }} orbit so

amp;12.75=13.6[112−1n2]amp;⇒n=4amp; So, Angular momentum L=nh2π=2hπ\begin{aligned}& 12.75=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right] \\\\& \Rightarrow n=4 \\\\& \text { So, Angular momentum } L=\frac{n h}{2 \pi}=\frac{2 h}{\pi} \end{aligned}

 Angular momentum =2πamp;×4.14×10−15amp;=828×10−17πeVs\begin{aligned}\text { Angular momentum }=\frac{2}{\pi} & \times 4.14 \times 10^{-15} \\\\& =\frac{828 \times 10^{-17}}{\pi} \mathrm{eVs}\end{aligned}

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